Question

A `2 kg` object is slide along a rough floor (coefficient of sliding friction `= 0.3)`with initial volocity `(10 m//s) hat i`. Describe the motion of the object relation to car assuming that the coefficient of static friction is greater than `0.5`.

Answer 1

Relative to car (non-inertial)

`a_(1)` is due to pseudo force
`a_(x) = - (5 + 3) = - 8 m//s^(2)`
Block will stop when
`v_(x) = 0 = v_(x) + a_(x)t = 10 - 8 t`
`t = 1.25 s`
So, for `t le 1.25 s`
`x = x_(0) + u_(x)t + (1)/(2) a_(x)t^(2)`
`= x_(0) + 10t - 4t^(2)`
`v_(x) = u_(x) + a_(x)t = 10 - 8t`
After this `mu _(s) g gt 5 m//s^(2)` as `mu _(s) gt0.5`
Therefore ,now the block remains stationary with respect to car.